Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 4 - Section 4.9 - Antiderivatives - 4.9 Exercises - Page 362: 21

Answer

$F\left( \theta \right) = - 2\cos \theta - 3\sec \theta + C$

Work Step by Step

$$\eqalign{ & f\left( \theta \right) = 2\sin \theta - 3\sec \theta \tan \theta \cr & {\text{Using the formulas in Table 2 }}\left( {{\text{see page 358}}} \right){\text{ }} \cr & {\text{Function: }}cf\left( x \right) \to {\text{Particular antiderivative: }}cF\left( x \right) \cr & {\text{Function: }}\sin x \to {\text{Particular antiderivative: }} - \cos x \cr & {\text{Function: }}\sec x\tan x \to {\text{Particular antiderivative: }}\sec x \cr & {\text{Function: }}f\left( x \right) + h\left( x \right) \to {\text{Particular antiderivative: }}f\left( x \right) + h\left( x \right) \cr & {\text{And applying the Theorem 1}},{\text{ we obtain}} \cr & f\left( \theta \right) = 2\sin \theta - 3\sec \theta \tan \theta \cr & F\left( \theta \right) = 2\left( { - \cos \theta } \right) - 3\left( {\sec \theta } \right) + C \cr & {\text{Simplifying}} \cr & F\left( \theta \right) = - 2\cos \theta - 3\sec \theta + C \cr & \cr & {\text{Checking the answer by differentiation}} \cr & F'\left( \theta \right) = \frac{d}{{d\theta }}\left[ { - 2\cos \theta } \right] - \frac{d}{{d\theta }}\left[ {3\sec \theta } \right] + \frac{d}{{dx}}\left[ C \right] \cr & F'\left( \theta \right) = - 2\left( { - \sin \theta } \right) - 3\left( {\sec \theta \tan \theta } \right) + 0 \cr & F'\left( \theta \right) = 2\sin \theta - 3\sec \theta \tan \theta \cr} $$
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